Show that $\int_{0}^{1} \phi(yt)\, dy$ is a characteristic function, where $\phi(t)$ is itself a characteristic function.

Question

I tried:

\begin{align}\int_{0}^{1} \phi(yt)\, dy &= \int_{0}^{1} E(e^{iytX})\, dy\\ &= \int_{0}^{1} E(\cos ytX + i\sin ytX)\, dy\\ &= \int_{0}^{1} E(\cos ytX)\, dy + i\int_{0}^{1} E(\sin ytX)\, dy\\ &= \text{??} \end{align}

No idea where to go from here.

Answer 1

Use Bochner's theorem which states a function is a characteristic function if and only if it is

1. positive semi-definite
2. The value at $0$ is $1$

So let be $$F(t) = \int_{0}^{1} \phi(yt)\, dy$$ then F is positive semi-definite because $\phi$ is and obviously $$F(0) = \int_{0}^{1} \phi(y\cdot0)\, dy = \int_{0}^{1} 1\, dy = 1$$

Hence $F$ is a characteristic function.

Answer 2

This is just a matter of translating between expectations and integrals, which are basically the same thing.

In probabilistic terms it's quite simple - just introduce an independent $\mathrm{Uniform}(0,1)$ variable $Y$, and note that $E(e^{itXY})=\int_0^1 E(e^{ityX})dy.$

In measure-theoretic terms, a function $\psi$ is a characteristic function iff it can be expressed as $$\psi(t)=\int_{\Omega} e^{itZ(\omega)} d\omega\tag{*}$$ for some measure space $\Omega$ of measure $1$ and some measurable function $Z:\Omega\to\mathbb R.$ You are given some measurable $X:\Omega_1\to\mathbb R$ and you need to construct $\Omega$ and $Z$ such that

$$\int_{0}^1 \int_{\Omega_1} e^{ityX(\omega_1)} d\omega dy=\int_{\Omega}e^{it Z(\omega)} d\omega$$

In these terms the answer is to take $\Omega=\Omega_1\times [0,1]$ and $Z(\omega_1,y)=X(\omega_1)y.$